# CBSE 10th Mathematics | Similar Triangles | Introduction

## Similar Triangles | Introduction

Similar figures: Geometric figures which have the same shape but different sizes are known as similar figures.

Illustrations:

1.    Any two line-segments are similar

2.    Any two squares are similar

3.    Any two circles are similar

Two congruent figures are always similar but two similar figures need not be congruent.

Similar polygons: Two polygons of the same number of sides are said to be similar if

(i)    Their corresponding angles are equal (i.e., they are equiangular) and

(ii)    Their corresponding sides are in the same ratio (or proportion)

Similar triangles: Since triangles are also polygons, the same conditions of similarity are applicable to them.

Two triangles are said to be similar if

(i)    Their corresponding angles are equal and

(ii)    Their corresponding sides are in the same ratio (or proportion).

## BASIC-PROPORTIONALITY THEOREM (Thales theorem)

Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively

To prove: $\displaystyle \frac{AD}{DB}=\frac{AE}{EC}.$

Construction: Join BE and CD and draw DM AC and EN AB.

Proof:     area of DADE $\displaystyle (=\frac{1}{2}\text{base}\times \text{height})=\frac{1}{2}AD\times EN.$

So,

$\displaystyle \text{ar}(ADE)=\frac{1}{2}AD\times EN$ [The area of DADE is denoted as ar (ADE)]

Similarly,

$\displaystyle \text{ar}(BDE)=\frac{1}{2}DB\times EN,$

$\displaystyle \text{ar}(ADE)=\frac{1}{2}AE\times DM$ and $\displaystyle \text{ar}(DEC)=\frac{1}{2}EC\times DM.$

(Taking AE as base)

Therefore,

$\displaystyle \frac{\text{ar}(ADE)}{\text{ar(}BDE\text{)}}=\frac{\frac{1}{2}AD\times EN}{\frac{1}{2}DB\times EN}=\frac{AD}{DB}$        …(i)

and

$\displaystyle \frac{\text{ar}(ADE)}{\text{ar(}DEC\text{)}}=\frac{\frac{1}{2}AE\times DM}{\frac{1}{2}EC\times DM}=\frac{AE}{EC}$            …(ii)

$\displaystyle \text{ar}(BDE)=\text{ar}(DEC)$                … (iii)

DBDE and DEC are on the same base DE and between the same parallels BC and DE.]

Therefore, from (i), (ii) and (iii), we have: $\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$

Corollary:

From above equation we have $\displaystyle \frac{DB}{AD}=\frac{EC}{AE}$

Adding ‘1’ to both sides we have

$\displaystyle \frac{DB}{AD}+1=\frac{EC}{AE}+1$

$\displaystyle \frac{DB+AD}{AD}=\frac{EC+AE}{AE}$ Þ $\displaystyle \frac{AB}{AD}=\frac{AC}{AE}$

Theorem: (Converse of BPT theorem) If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side.

Given:

In DABC, DE is a straight line such that $\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$

To prove:

DE || BC

Construction:

If DE is not parallel to BC, draw DF meeting AC at F

Proof:

In DABC, let DF || BC

\    $\displaystyle \frac{AD}{DB}=\frac{AF}{FC}$                        …(i)

[$\displaystyle \because$ A line drawn parallel to one side of a D divides the other two sides in the same ratio.]

But    $\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$                        …(ii)    [given]

From (i) and (ii), we get

$\displaystyle \frac{AF}{FC}=\frac{AE}{EC}$

Adding 1 to both sides, we get

$\displaystyle \frac{AF}{FC}+1=\frac{AE}{EC}+1$

Þ        $\displaystyle \frac{AF+FC}{FC}=\frac{AE+EC}{EC}$

Þ        $\displaystyle \frac{AC}{FC}=\frac{AC}{EC}$ Þ
FC = EC

It is possible only when E and F coincide

Hence,     DE || BC.

### Solution:

It has been given that

PQ = 15.2 cm, PR = 12.8 cm,

PM = 5.7 cm and PN = 4.8 cm

\     MQ = PQPM = (15.2 – 5.7) cm = 9.5 cm and

NR = PR – PN = (12.8 – 4.8) cm = 8 cm

Now

$\displaystyle \frac{PM}{MQ}=\frac{5.7}{9.5}=\frac{3}{5}$ and $\displaystyle \frac{PN}{NR}=\frac{4.8}{8}=\frac{3}{5}$

\    $\displaystyle \frac{PM}{MQ}=\frac{PN}{NR}$

Thus, in DPQR, MN divides the sides PQ and PR in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem, we have MN II QR.

### Solution:

Since the diagonals of a trapezium divide each other proportionally

\    $\displaystyle \frac{AO}{OC}=\frac{BO}{OD}$

Þ    $\displaystyle \frac{7x-9}{2x-1}=\frac{9x-8}{3x}$

Þ    $\displaystyle 21{{x}^{2}}-27x=18{{x}^{2}}-9x-16x+8$

Þ    $\displaystyle 21{{x}^{2}}-27x=18{{x}^{2}}-25x+8$

Þ    $\displaystyle 3{{x}^{2}}-2x-8=0$

Þ    $\displaystyle 3{{x}^{2}}-6x+4x-8=0$

Þ    3x(x – 2) + 4 (x – 2) = 0

Þ    $\displaystyle (x-2)\,(3x+4)=0$

Þ    Either $\displaystyle x-2=0$    or    $\displaystyle 3x+4=0$

Þ    $\displaystyle x=2$            or    $\displaystyle x=-\frac{4}{3}$

$\displaystyle [x=-\frac{4}{3},$ rejected, as it makes line-segments negative]

Thus, $\displaystyle x=2.$

### Solution:

(i)

$\displaystyle \because$ In DABC, DE || BC then by Basic Proportionality (BPT) Theorem, we have

$\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$    or     cm

(ii)

Also in figure (ii), in DABC, DE || BC then by B.P.T, we have

$\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$    or     $\displaystyle AD=\frac{AE\times DB}{EC}=\frac{1.8\times 7.2}{5.4}$ = 2.4 cm

### Solution:

Given:

LM || CB and LN || CD

To prove:

$\displaystyle \frac{AM}{AB}=\frac{AN}{AD}$

Proof:

In DABC, LM || BC, then

By BPT Theorem, we have

$\displaystyle \frac{AM}{AB}=\frac{AL}{AC}$                …(i)     [Corollary of B.P.T.]

Similarly, in DADC, LN || CD

$\displaystyle \frac{AN}{AD}=\frac{AL}{AC}$                …(ii)    [Corollary of B.P.T.]

From (i) and (ii), we have $\displaystyle \frac{AM}{AB}=\frac{AN}{AD}$

### Solution:

In DABC, DE || AC then by B.P.T., we have

$\displaystyle \frac{BD}{DA}=\frac{BE}{EC}$                    …(i)

In DABE, DF || AE then by B.P.T., we have

$\displaystyle \frac{BD}{DA}=\frac{BF}{EF}$                    …(ii)

From (i) and (ii), we get

$\displaystyle \frac{BF}{EF}=\frac{BE}{CE}$

### Solution:

In DPOQ, DE || OQ, then by BPT Theorem, we have

$\displaystyle \frac{PE}{EQ}=\frac{PD}{DO}$                    …(i)

Also, in DPOR, DF || OR, then by BPT Theorem, we have

$\displaystyle \frac{PF}{FR}=\frac{PD}{DO}$     …(ii)

From equations, (i) and (ii),

$\displaystyle \frac{PE}{EQ}=\frac{PF}{FR}$

Þ    $\displaystyle EF||QR$        [converse of BPT Theorem]

### Solution:

In DPOQ, PQ || AB, then by B.P.T., we have

$\displaystyle \frac{OP}{PA}=\frac{OQ}{QB}$    …(i)

Also in DPOR, AC || PR, we have

$\displaystyle \frac{OP}{PA}=\frac{OR}{RC}$                    …(ii)

From (i) and (ii), we get

$\displaystyle \frac{OQ}{QB}=\frac{OR}{RC}$

Þ    QR || BC             [by converse of B.P.T.]

### Solution:

Given: A DABC in which D is the mid point of AB and DE || BC meeting AC at E.

To prove:
AE = CE.

Proof:     In DABC, DE || BC

\    $\displaystyle \frac{AD}{BD}=\frac{AE}{CE}$            [BPT Theorem]

But    AD = BD             [D is the mid point of AB]

\    $\displaystyle \frac{AE}{CE}=1$    Þ    AE = CE

Hence, E is the mid point of AC.

### Solution:

Given:

ABCD is a trapezium such that AB || DC

To prove:

$\displaystyle \frac{AO}{OC}=\frac{BO}{DO}$ or $\displaystyle \frac{AO}{BO}=\frac{CO}{DO}$

Construction:

Through O draw OE || CD.

Proof:

AB || DC    [given]

and         EO || DC    [const.]

Þ        EO || AB

($\displaystyle \because$ Two lines parallel to same line are parallel to each other)

Now, in DABD, EO || AB, then by BPT Theorem we have

$\displaystyle \frac{AE}{ED}=\frac{BO}{DO}$                …(i)

Also in DADC, EO || DC, by BPT Theorem, we have

$\displaystyle \frac{AE}{ED}=\frac{AO}{OC}$                …(ii)

From (i) and (ii), we get

$\displaystyle \frac{BO}{DO}=\frac{AO}{OC}$    or    $\displaystyle \frac{AO}{BO}=\frac{CO}{DO}$

### Solution:

Given:

A quadrilateral ABCD, whose diagonals AC and BD intersect at O such that

$\displaystyle \frac{AO}{BO}=\frac{CO}{DO}$    or    $\displaystyle \frac{AO}{OC}=\frac{BO}{OD}$

To prove:

ABCD is a trapezium.

Construction:

Though O, draw a line OE parallel to AB intersecting BC at E.

Proof:

In DABC, OE || AB     [By construction]

\    $\displaystyle \frac{AO}{OC}=\frac{BE}{EC}$             [by BPT Theorem]    …(i)

$\displaystyle \frac{AO}{OC}=\frac{BO}{OD}$             [given]            …(ii)

From (i) and (ii), we have $\displaystyle \frac{BO}{OD}=\frac{BE}{EC}$

Now, in DDBC, we have     $\displaystyle \frac{BO}{OD}=\frac{BE}{EC}$

Þ    OE || DC         [Converse of BPT Theorem]

Now, in quadrilateral ABCD, OE || AB and OE || DC
Þ
AB || DC

($\displaystyle \because$ Two lines parallel to same line are parallel to each other)

\    Quadrilateral ABCD is a trapezium.